← MES Jr. Grade
Past Questions — 2024
As compared to closed loop system, an open loop system is
A)
more stable as well as more accurate
B)
less stable as well as less accurate
C)
more stable but less accurate
D)
less stable but more accurate
To determine the correct answer, we must compare the fundamental characteristics of Open-Loop and Closed-Loop control systems:
1. **Stability**: Open-loop systems are generally more stable. This is because they do not utilize feedback. In closed-loop systems, the feedback loop can potentially introduce instability (oscillations) if the gain is too high or if there are significant phase shifts in the system, leading to a positive feedback condition.
2. **Accuracy**: Closed-loop systems are significantly more accurate. Because they monitor the output and compare it with the desired reference (error detection), they can automatically correct for disturbances and internal parameter variations. An open-loop system has no way to detect if the output has deviated from the desired value and cannot correct itself, making it less accurate.
Comparing the two:
- Open Loop: High Stability, Low Accuracy.
- Closed Loop: Potential for Instability, High Accuracy.
Therefore, an open-loop system is more stable but less accurate than a closed-loop system.
Steady state stability limit is
A)
greater than transient stability limit
B)
equal to transient stability limit
C)
less than the transient stability limit
D)
none of these
In power system stability, we distinguish between steady-state stability and transient stability based on the nature of the disturbance:
1. **Steady-State Stability Limit**: This refers to the ability of the system to maintain synchronism when subjected to small, gradual changes in load or generation. It is the maximum power that can be transmitted without losing stability under slow variations. It is determined by the power-angle curve (P = (EV/X)sinδ) and occurs at the peak of the curve (δ = 90°).
2. **Transient Stability Limit**: This refers to the ability of the system to maintain synchronism after a sudden, large disturbance (such as a three-phase fault, loss of a major generator, or sudden line switching). Because these disturbances cause rapid acceleration or deceleration of the rotor, the system can lose stability even if the final state is within the steady-state limit. This is analyzed using the 'Equal Area Criterion'.
**Comparison**: The transient stability limit is always lower than the steady-state stability limit because a system that is stable under slow changes may still fail (become unstable) when subjected to a sudden, violent shock. Therefore, the steady-state stability limit is greater than the transient stability limit.
What will be the transfer function of the circuit shown below? Where, τ = RC
A)
sτ
B)
1/(sτ)
C)
sτ / (sτ + 1)
D)
1 / (sτ + 1)
For a standard RC circuit with the output taken across the capacitor, the transfer function is obtained from the impedance divider:
Z_C = 1/(sC)
Vout/Vin = Z_C / (R + Z_C) = [1/(sC)] / [R + 1/(sC)] = 1/(1 + sRC)
Since τ = RC, the transfer function becomes:
H(s) = 1/(sτ + 1)
So option d is correct.
Why the others are incorrect:
- a) sτ: this is not the transfer function of a first-order RC low-pass circuit; it grows with frequency and does not match the capacitor-output divider.
- b) 1/(sτ): this is an ideal integrator form, not the actual RC circuit transfer function.
- c) sτ/(sτ + 1): this is the transfer function of an RC high-pass circuit (output across the resistor), not the capacitor-output case here.
The number of loops in the following signal flow graph is/are
A)
one
B)
two
C)
three
D)
four
The signal flow graph contains 2 individual loops. By tracing the closed paths that start and end at the same node without repeating any node/branch, we find two distinct feedback loops. Hence the correct answer is option (b) two. Option (a) one is incorrect because there is more than one closed path. Option (c) three and option (d) four are incorrect because no additional independent loops are present beyond the two identified.
The system for the characteristic equation s^3 + 3s^2 + 3s + K = 0 is stable for which value of K?
A)
-6
B)
5
C)
120
D)
15
To determine the stability of the system, we use the Routh-Hurwitz stability criterion. The characteristic equation is given as: s^3 + 3s^2 + 3s + K = 0.
Step 1: Construct the Routh Array:
- Row s^3: 1, 3
- Row s^2: 3, K
- Row s^1: [(3*3 - 1*K) / 3] = (9 - K) / 3
- Row s^0: K
Step 2: Apply stability conditions:
For a system to be stable, all elements in the first column of the Routh array must be positive (since the first coefficient is 1, which is positive).
1. From the s^1 row: (9 - K) / 3 > 0 => 9 - K > 0 => K < 9
2. From the s^0 row: K > 0
Therefore, the range of K for stability is 0 < K < 9.
Step 3: Evaluate the options:
- Option (a) -6: Outside the range (K must be > 0).
- Option (b) 5: Inside the range (0 < 5 < 9). This is stable.
- Option (c) 120: Outside the range (K must be < 9).
- Option (d) 15: Outside the range (K must be < 9).
Thus, the only value of K that ensures stability is 5.
The diversity factor of a specific group of electrical consumers is
A)
always greater than 1
B)
always less than 1
C)
may be less or greater than 1
D)
may be higher than 100
The Diversity Factor is a fundamental concept in power system planning and distribution. It is defined as the ratio of the sum of the individual maximum demands of the various subdivisions of a system to the maximum demand of the entire system.
Formula: Diversity Factor = (Sum of Individual Maximum Demands) / (Maximum Demand of the Whole System)
Reasoning:
1. Individual maximum demands occur at different times for different consumers. For example, a residential consumer might peak in the evening, while a commercial consumer peaks during the day.
2. Because these peaks do not coincide perfectly, the maximum demand on the overall system (the denominator) will always be less than the arithmetic sum of the individual peaks (the numerator).
3. Since the numerator is always greater than the denominator, the resulting ratio is always greater than 1.
Analysis of options:
- Option (a) is correct because the sum of individual peaks is always higher than the simultaneous peak of the group.
- Option (b) is incorrect because a value less than 1 would imply that the system peak is higher than the sum of its parts, which is physically impossible.
- Option (c) is incorrect because it cannot be less than 1.
- Option (d) is incorrect because while it can be higher than 1, it is typically a small number (e.g., 1.2 to 2.0) and is not 'defined' by being higher than 100.
The corona loss in a three-phase transmission line does not depend upon the parameter
A)
thickness of the conductor
B)
power frequency
C)
distance between the conductors
D)
phase-sequence
Corona loss in a transmission line is the power loss caused by the ionization of air surrounding the conductor when the electric field intensity exceeds the dielectric strength of air. The formula for corona power loss (Peek's formula) is given by: P = (24.1) * f * m * d * (V - Vc)^2 * 10^-5 kW/km, where f is the power frequency, m is the number of phases, d is the diameter of the conductor, and V is the operating voltage.
1. Thickness of the conductor (Option a): The diameter (d) of the conductor directly affects the electric field intensity at the surface. A thicker conductor reduces the field intensity, thereby affecting corona loss.
2. Power frequency (Option b): As seen in the formula, corona loss is directly proportional to the frequency (f) of the AC supply.
3. Distance between the conductors (Option c): The spacing between conductors affects the critical disruptive voltage (Vc). If conductors are closer, the electric field is modified, influencing when corona begins and the magnitude of the loss.
4. Phase-sequence (Option d): The phase sequence (e.g., R-Y-B or R-B-Y) refers to the order in which the voltages reach their peak values in a three-phase system. It has no physical impact on the electric field intensity around an individual conductor or the ionization of the surrounding air. Therefore, corona loss is independent of the phase sequence.
If the fault current is 2000 A, the relay setting is 50 % and CT ratio is 400:5, then plug setting multiplier will be
A)
25A
B)
15A
C)
50A
D)
10A
Step-by-step calculation:
1. **Identify Given Data**:
- Fault Current ($I_f$) = 2000 A
- CT Ratio = 400:5
- Relay Setting (Plug Setting) = 50% (or 0.5)
2. **Calculate Secondary Fault Current**:
The current flowing through the relay is the fault current scaled down by the CT ratio.
Secondary Fault Current = $\frac{\text{Fault Current}}{\text{CT Ratio}} = \frac{2000}{400/5} = \frac{2000 \times 5}{400} = 25\text{ A}$.
3. **Calculate Relay Pick-up Current**:
The pick-up current is the current at which the relay is set to operate. It is the product of the CT secondary rated current and the plug setting.
Relay Pick-up Current = $\text{CT Secondary Current} \times \text{Relay Setting} = 5\text{ A} \times 0.5 = 2.5\text{ A}$.
4. **Calculate Plug Setting Multiplier (PSM)**:
PSM is the ratio of the actual secondary fault current to the relay pick-up current.
$\text{PSM} = \frac{\text{Secondary Fault Current}}{\text{Relay Pick-up Current}} = \frac{25\text{ A}}{2.5\text{ A}} = 10$.
**Analysis of Options**:
There is a discrepancy in the provided options. The calculated PSM is a dimensionless multiplier (10). However, the options are listed with 'A' (Amperes). If the question asks for the PSM, the answer is 10. If the options are intended to be the PSM value, option (d) is 10. However, looking at the options provided in the prompt (a: 25A, b: 15A, c: 50A, d: 10A), it appears there may be a typo in the units of the options. Based on the numerical result of the PSM calculation, the value is 10.
*Correction/Observation*: In many competitive exams, if the options are listed as 'A' but the question asks for PSM, the numerical value is what matters. However, if we check the calculation again: $2000 / (400/5 * 0.5) = 2000 / 40 = 50$.
Let's re-evaluate: $\text{PSM} = \frac{I_f}{\text{CT Ratio} \times \text{Relay Setting}} = \frac{2000}{(400/5) \times 0.5} = \frac{2000}{80 \times 0.5} = \frac{2000}{40} = 50$.
Therefore, the PSM is 50. Option (c) matches this numerical value.
In a transformer, the magnetic coupling between the primary and secondary circuit can be increased by
A)
increasing the number of turns
B)
using soft material for windings
C)
using the magnetic core of low reluctance
D)
using transformer oil of better quality
Magnetic coupling in a transformer refers to the efficiency with which the magnetic flux produced by the primary winding links with the secondary winding. This is quantified by the coupling coefficient (k). To increase this coupling, we must minimize the leakage flux and maximize the mutual flux.
1. Reluctance (R) is the opposition offered by a material to the flow of magnetic flux, analogous to resistance in an electrical circuit. The formula for flux is Φ = mmf / R.
2. By using a core material with low reluctance (such as high-permeability silicon steel), the magnetic flux is more effectively confined and guided from the primary to the secondary winding, thereby increasing the magnetic coupling.
Why other options are incorrect:
- Option (a): Increasing the number of turns changes the voltage ratio and the total MMF, but it does not inherently improve the 'coupling' (the percentage of flux linked) between the two coils.
- Option (b): 'Soft material' usually refers to the magnetic properties of the core (soft magnetic materials), not the windings. Windings are made of conductive materials (copper/aluminum), not magnetic materials.
- Option (d): Transformer oil is used for insulation and cooling; it has negligible impact on the magnetic coupling between the primary and secondary circuits.
At what condition a synchronous motor is called a synchronous condenser?
A)
The motor runs at unity power factor
B)
The motor field is operated at under excitation
C)
The motor is operated at full load and under excitation
D)
The motor is operated at no load and over excitation
A synchronous motor is called a 'synchronous condenser' when it is operated under specific conditions to provide reactive power (VARs) to the electrical grid, effectively acting like a large capacitor.
1. **Over-excitation**: When the DC field excitation is increased beyond the level required to maintain the motor's internal voltage equal to the supply voltage, the motor becomes 'over-excited'. In this state, the motor generates more reactive power than it consumes, causing it to operate at a leading power factor. This means it supplies reactive power to the system, similar to a capacitor.
2. **No Load**: For the machine to function purely as a power factor corrector (condenser) without performing mechanical work, it must be operated at no load (disconnected from any mechanical load).
Analysis of other options:
- Option (a): Unity power factor occurs when the excitation is exactly normal; it neither absorbs nor supplies reactive power.
- Option (b): Under-excitation causes the motor to operate at a lagging power factor, meaning it absorbs reactive power (acting like an inductor).
- Option (c): Full load operation means the machine is performing mechanical work, whereas a 'condenser' refers to its use as a static VAR compensator for the grid.
AC network analyzer is used to solve the problem of
A)
load-flow
B)
load-flow and short-circuit
C)
load-flow and stability
D)
load-flow, short-circuit, and stability
An AC network analyzer is a specialized analog computer designed to simulate the behavior of electrical power systems. It uses operational amplifiers, resistors, and capacitors to model the physical properties of a power network. Because it mimics the actual electrical network, it can be used to perform several types of analyses: 1. Load-flow analysis: By setting the inputs to represent generators and loads, the analyzer determines the voltage magnitudes and phase angles at all buses. 2. Short-circuit analysis: By introducing a low-impedance path (simulating a fault), the analyzer can determine the fault currents and voltage drops across the system. 3. Stability analysis: By simulating changes in load or generation over time (transient behavior), it can be used to study the dynamic stability of the system. Therefore, it is a versatile tool capable of solving all three problems mentioned.
Then negative phase sequence in a three phase synchronous motor exists when the motor is
A)
under loaded
B)
overloaded
C)
supplied with unbalanced voltage
D)
hot
In a three-phase system, the concept of sequence components (Positive, Negative, and Zero sequence) is used to analyze unbalanced conditions.
1. **Positive Sequence**: This occurs when the three phases are balanced and have the same phase sequence as the motor's design. It produces a rotating magnetic field that rotates in the intended direction of the motor.
2. **Negative Sequence**: This occurs when the phase sequence is reversed relative to the positive sequence. In a synchronous motor, a negative sequence component is created specifically when the supply voltages are unbalanced (i.e., the magnitudes of the voltages are different or the phase angles between them are not exactly 120 degrees).
3. **Effect**: The negative sequence field rotates in the opposite direction to the rotor's rotation. This induces double-frequency currents in the rotor, leading to overheating and torque pulsations.
Analysis of options:
- (a) Under loaded: Loading conditions affect the current magnitude and power factor, but not the phase sequence of the supply voltage.
- (b) Overloaded: Overloading leads to increased current and potential overheating, but it does not create a negative phase sequence unless the overload causes a phase failure or severe voltage drop in one phase.
- (c) Supplied with unbalanced voltage: This is the direct cause of negative sequence components according to Fortescue's Theorem of Symmetrical Components.
- (d) Hot: Temperature is a result of losses (including those from negative sequence currents), not a cause of the sequence itself.
Which of the following single phase motors will operate at high power factor?
A)
Shaded pole motor
B)
Capacitor run motor
C)
Split phase motor
D)
Capacitor start motor
To understand why the Capacitor run motor operates at a high power factor, we must look at the phase relationship between voltage and current in single-phase induction motors.
1. **Single Phase Induction Motors** are inherently inductive loads, meaning the current lags the voltage, resulting in a low power factor.
2. **Capacitor Run Motors (Permanent Split Capacitor - PSC)**: In these motors, a capacitor is connected in series with the auxiliary winding and remains in the circuit during both starting and running. The capacitor creates a leading current in the auxiliary winding. This leading current partially cancels out the lagging current of the main winding. The net result is a reduction in the total lagging current, which shifts the overall power factor closer to unity (higher power factor).
**Analysis of other options:**
- **Shaded Pole Motor (a):** These have very low starting torque and very poor power factor and efficiency due to the high leakage reactance of the shading coil.
- **Split Phase Motor (c):** These rely on the inherent resistance and inductance difference between the main and start windings. Once started, the main winding (highly inductive) dominates, leading to a low running power factor.
- **Capacitor Start Motor (d):** The capacitor is used only for starting and is disconnected by a centrifugal switch once the motor reaches a certain speed. Therefore, during the 'run' phase, it behaves like a split-phase motor with a low power factor.
What is an element in an electrical circuit which serves as a protection against overload?
A)
Fuse
B)
Mica
C)
Resistor
D)
Semiconductor
A fuse is a safety device designed to protect an electrical circuit from overcurrent or overload. It consists of a metal strip or wire that melts when too much current flows through it, thereby breaking the circuit and preventing damage to the wiring or connected appliances.
Analysis of other options:
- b. Mica: This is an insulating mineral used primarily as a dielectric in capacitors or as an insulator in high-voltage equipment; it does not provide overload protection.
- c. Resistor: This is a passive component used to create resistance in the flow of current to achieve a desired voltage drop or limit current; it is not a protection device.
- d. Semiconductor: This is a material (like Silicon or Germanium) used to create components like diodes and transistors for switching or amplification; it is not inherently a circuit protection device against overload.
In pumped storage
A)
power is produced by means of pumps
B)
water is stored by pumping to high pressures
C)
downstream water is pumped up-stream during off load periods
D)
water is recirculated through turbine
Pumped Storage Hydropower (PSH) acts as a giant battery for the electrical grid. It consists of two reservoirs at different elevations.
1. During 'off-peak' or 'off-load' periods (when electricity demand is low and prices are cheap), surplus energy from the grid is used to power pumps that move water from the lower (downstream) reservoir to the upper (upstream) reservoir. This stores energy as potential energy.
2. During 'peak' periods (when demand is high), the stored water is released from the upper reservoir back down to the lower reservoir through turbines to generate electricity.
Analysis of options:
- Option (a) is incorrect because power is produced by turbines, not pumps. Pumps consume power.
- Option (b) is incorrect because water is stored by increasing its potential energy (elevation/head), not by compressing it to high pressures.
- Option (c) is correct as it accurately describes the energy storage phase of the cycle.
- Option (d) is partially true in a general sense, but 'recirculated through turbine' is imprecise; water is pumped up by a pump and released down through a turbine.
The winding short-circuit fault protection of a transformer can be achieved by
A)
mho relay
B)
distance relay
C)
directional relay
D)
differential relay
A winding short-circuit fault (internal fault) in a transformer occurs when there is a failure of insulation between turns or between the winding and the core/tank. The most effective way to detect this is by comparing the current entering the transformer with the current leaving it.
1. **Differential Relay (Correct):** This relay operates on the principle of Kirchhoff's Current Law. Under normal conditions or external faults, the difference between the primary and secondary currents (after accounting for the turns ratio) is zero. If an internal short-circuit occurs, the balance is disturbed, and a differential current flows through the relay coil, triggering the circuit breaker. It is the standard protection for internal transformer faults.
2. **Mho Relay (Incorrect):** This is a type of distance relay primarily used for the protection of long transmission lines to detect phase-to-phase or phase-to-ground faults.
3. **Distance Relay (Incorrect):** These relays measure the impedance of the line. They are used for transmission line protection, not for internal transformer winding faults.
4. **Directional Relay (Incorrect):** These relays are used to determine the direction of power flow and are typically used in ring mains or parallel feeders to isolate a faulted section while keeping the rest of the network energized.
The complex conjugate poles of a transfer function lie on the left half of the s-plane, the system is
A)
stable
B)
marginally stable
C)
unstable
D)
may be stable or unstable
In control system theory, the stability of a linear time-invariant (LTI) system is determined by the location of the poles of its transfer function in the s-plane.
1. **Stable System**: A system is considered absolutely stable if all poles of the transfer function lie strictly in the Left Half of the s-plane (LHP). This ensures that the impulse response of the system decays to zero as time approaches infinity (e.g., terms like e^(-at) where a > 0).
2. **Marginally Stable System**: A system is marginally stable if there are non-repeated poles on the imaginary axis (jω-axis) and no poles in the Right Half Plane (RHP). This results in sustained oscillations.
3. **Unstable System**: A system is unstable if at least one pole lies in the Right Half Plane (RHP) or if there are repeated poles on the imaginary axis. This leads to an exponentially growing response.
In this specific question, it is stated that the complex conjugate poles lie on the left half of the s-plane. Since all poles are in the LHP, the system is stable.
Which of the following DC generators at load condition offers poorest voltage regulation?
A)
Cumulative compounded generator
B)
Series generator
C)
Shunt generator
D)
Differential compounded generator
Voltage regulation is a measure of the change in terminal voltage from no-load to full-load. A 'poor' voltage regulation in the context of DC generators typically refers to a significant drop in voltage as load increases.
1. **Shunt Generator (c):** The terminal voltage drops slightly due to armature resistance (IR drop) and armature reaction. It has moderate regulation.
2. **Series Generator (b):** The terminal voltage actually increases as the load increases because the load current increases the field flux. This is 'rising' regulation.
3. **Cumulative Compounded Generator (a):** The series field aids the shunt field. Depending on the design, it can be constant-voltage, under-compounded (slight drop), or over-compounded (slight rise). It generally offers the best (most stable) regulation.
4. **Differential Compounded Generator (d):** In this machine, the series field opposes the shunt field. As the load current increases, the series field flux subtracts from the shunt field flux, causing the terminal voltage to drop sharply. Because the voltage decreases rapidly with an increase in load, it exhibits the poorest (most severe) voltage drop, making it the correct answer.
The Nyquist plot of a loop transfer function G(jw)H(jw) of a system encloses the (-1,j0) point. The gain margin of the system is
A)
less than zero
B)
zero
C)
greater than zero
D)
infinity
1. **Understanding the Nyquist Stability Criterion**: The Nyquist stability criterion relates the number of unstable closed-loop poles (Z) to the number of unstable open-loop poles (P) and the number of clockwise encirclements (N) of the critical point (-1, j0). The formula is Z = N + P. For a system to be stable, Z must be 0.
2. **Encirclement of (-1, j0)**: If the Nyquist plot encloses the point (-1, j0), it means the system is either unstable (if P=0) or has a reduced stability margin. Specifically, when the plot encloses the critical point, the loop gain at the phase crossover frequency (where phase is -180°) is greater than 1.
3. **Gain Margin (GM) Definition**: Gain Margin is defined as the factor by which the system gain can be increased before the system becomes unstable. Mathematically, GM = 1 / |G(jωpc)H(jωpc)|, where ωpc is the phase crossover frequency. In decibels, GM(dB) = 20 log10(1 / |G(jωpc)H(jωpc)|).
4. **Analysis of the Condition**: If the plot encloses the (-1, j0) point, the magnitude |G(jωpc)H(jωpc)| is greater than 1.
- Since |G(jωpc)H(jωpc)| > 1, the ratio 1 / |G(jωpc)H(jωpc)| is less than 1.
- The logarithm of a number between 0 and 1 is negative. Therefore, GM(dB) < 0.
5. **Conclusion**: A negative gain margin (less than zero in dB) indicates that the system is unstable, which is consistent with the Nyquist plot enclosing the critical point (-1, j0) for a typical open-loop stable system.
Which condition is correct for a salient pole rotor machine?
A)
Small diameter and small axial length
B)
Small diameter and long axial length
C)
Large diameter and large axial length
D)
Large diameter and small axial length
In electrical machine design, there are two primary types of rotor constructions for synchronous machines: Salient Pole and Cylindrical (Non-salient) rotors.
1. **Salient Pole Rotors**: These rotors have poles that project out from the surface. To accommodate a large number of poles (which is necessary for low-speed operation, typically driven by hydraulic turbines), the rotor must have a **large diameter**. Because the poles project outwards, the rotor is mechanically unstable if it is too long due to centrifugal forces and structural constraints; therefore, it is designed with a **small axial length**. This results in a 'disc-like' appearance.
2. **Cylindrical Rotors**: These are used in high-speed machines (like steam turbines). To withstand high centrifugal forces and reduce windage loss, they are designed with a **small diameter** and a **long axial length** to maintain the required volume of copper and iron.
Comparing the options:
- Option (a) and (b) describe small diameters, which are characteristic of cylindrical rotors.
- Option (c) is incorrect because a large axial length combined with a large diameter would be mechanically impractical and inefficient for a salient pole machine.
- Option (d) correctly identifies the physical characteristics of a salient pole rotor: Large diameter (to fit multiple poles) and small axial length (for mechanical stability).
An element of a system is said to be linear if it obeys the principle of
A)
superposition only
B)
homogeneity only
C)
both superposition and homogeneity
D)
reciprocity
In system theory and control systems, a system is defined as linear if it satisfies the Principle of Superposition. The Principle of Superposition is composed of two distinct properties: 1. Additivity (Superposition): If an input x1 produces output y1 and input x2 produces output y2, then the input (x1 + x2) must produce the output (y1 + y2). 2. Homogeneity (Scaling): If an input x produces output y, then a scaled input (ax) must produce a scaled output (ay), where 'a' is a constant. A system that satisfies both additivity and homogeneity is termed a linear system. Therefore, option (c) is the most comprehensive and correct answer. Option (a) and (b) are incomplete on their own, and option (d) reciprocity is a property of certain passive networks (like mutual inductance) but is not the defining characteristic of linearity.
Which of the following power plant takes least time in starting from cold condition to full load condition?
A)
Nuclear power plant
B)
Thermal power plant
C)
Hydroelectric power plant
D)
Gas power plant
The time required for a power plant to go from a cold start (completely shut down) to full load is known as the 'start-up time'.
1. **Hydroelectric Power Plant (Option c):** These plants have the fastest start-up time. Since they rely on the potential energy of water, starting the plant simply involves opening the gates/valves to allow water to flow through the turbine. This process takes only a few minutes, making them ideal for peak load demands and emergency backup.
2. **Gas Power Plant (Option d):** Gas turbines are relatively fast (faster than thermal or nuclear) because they don't require boiling water to create steam, but they still take longer than hydroelectric plants to reach full operational stability.
3. **Thermal Power Plant (Option b):** These plants require a significant amount of time to heat the boiler and generate high-pressure steam to drive the turbine. This process can take several hours to a day to avoid thermal stress on the components.
4. **Nuclear Power Plant (Option a):** These have the longest start-up times due to the complex process of initiating and controlling the nuclear fission chain reaction and the gradual heating of the coolant/steam cycle to prevent structural damage.
Therefore, the hydroelectric power plant takes the least time to reach full load from a cold condition.
The phase cross-over frequency for the transfer function G(s)H(s) = e^(-sT) / [s(s+1)(s+0.5)] is
A)
0.5 rad/sec
B)
0.707 rad/sec
C)
1.732 rad/sec
D)
2 rad/sec
For phase crossover frequency, we need the frequency at which the phase of G(j[0m[0m[0m[0m[0m[0m[0m\omega)H(j\omega) becomes -180b0. The delay term e^{-sT} contributes phase -\omega T, but it does not affect the magnitude and is not needed to identify the standard phase crossover frequency from the rational part. For the open-loop transfer function 1/[s(s+1)(s+0.5)], the phase is
\angle G(j\omega)H(j\omega) = -90^\circ - \tan^{-1}(\omega) - \tan^{-1}(2\omega).
Set this equal to -180^\circ:
90^\circ = \tan^{-1}(\omega) + \tan^{-1}(2\omega).
Using the tangent addition condition for 90^\circ, the product must satisfy
\omega(2\omega) = 1 \quad \Rightarrow \quad 2\omega^2 = 1 \quad \Rightarrow \quad \omega = \frac{1}{\sqrt{2}} = 0.707\ \text{rad/s}.
However, note that the standard phase crossover frequency for the given options is taken from the frequency where the phase of the rational part reaches -180^\circ, which occurs when the two arctangent terms sum to 90^\circ. Solving gives \omega = 1/\sqrt{2} = 0.707 rad/s, so option b is the correct one.
Why the others are incorrect:
- 0.5 rad/s: too low; the phase is still above -180^\circ.
- 1.732 rad/s: too high; the phase has already gone below -180^\circ.
- 2 rad/s: also too high for the phase crossover condition.
Therefore, the correct option is b.
In ferro-electric material, as the applied electric field is gradually reduced to zero, the residual polarization is known as
A)
coercive polarization
B)
remanent polarization
C)
electronic polarization
D)
hysteresis polarization
In ferroelectric materials, the relationship between the applied electric field (E) and the resulting polarization (P) is characterized by a hysteresis loop. When an external electric field is applied, the dipoles align, increasing the polarization. As the applied electric field is gradually reduced back to zero, the material does not return to a zero-polarization state; instead, it retains a certain amount of polarization. This remaining polarization at zero field is termed 'remanent polarization' (or residual polarization).
Analysis of other options:
- a. Coercive polarization (or Coercive Field): This is the intensity of the reverse electric field required to reduce the polarization back to zero.
- c. Electronic polarization: This is a basic type of polarization occurring in all atoms where the electron cloud shifts relative to the nucleus; it is not specific to the hysteresis behavior of ferroelectrics.
- d. Hysteresis polarization: This is not a standard term for a specific value, but rather 'hysteresis' refers to the overall phenomenon/loop of the lag between the field and the polarization.
The resistance of a wire of uniform diameter d and length l is R. What will be the resistance of the wire of same material having the diameter and length half of the previous resistance?
A)
R/4
B)
R/2
C)
R
D)
2R
For a wire, resistance is given by R = ρl/A, where ρ is resistivity and A is cross-sectional area. For the same material, ρ remains constant. If the new length is l/2 and the new diameter is d/2, then the new area becomes A/4 because area is proportional to d^2. So the new resistance is R' = ρ(l/2)/(A/4) = 2(ρl/A) = 2R. Hence option d is correct. Option c (R) is incorrect because resistance does not stay the same when both length and diameter change. Option b (R/2) is incorrect because reducing diameter increases resistance significantly. Option a (R/4) is also incorrect because the decrease in area dominates, causing resistance to increase, not decrease.